Re: How real are the "Virtual" partticles?

From: Eugene Stefanovich (eugenev_at_synopsys.com)
Date: 03/16/05 

Date: Wed, 16 Mar 2005 16:19:02 +0000 (UTC)

Igor Khavkine wrote:

> Can you give a reference to your approach besides the PDF file
> from your website?

In addition to the book www.meopemuk.com/book.pdf
there are two papers: Ann. Phys. 292 (2001), 139 and hep-th/0503076.
A couple of related papers can be also found on
  www.geocities.com/meopemuk

>> When you draw a real electron in QED you first draw
>>a line corresponding to a "bare" electron. Then you add lines of virtual
>>photons that begin and end on the bare line. Then you add small
>>virtual electron-positron loops to the virtual photon lines, etc.
>>So, loosely speeking, in QED real electron = bare electron + coat of
>>virtual particles.
>
>
> That is one way to look at it, but is not the only one. I need not draw
> bare particle lines at all. If I wish, I can only draw dressed lines.
> In this formulation scattering amplitudes can still be expanded in
> terms of Feynman diagrams, but only dressed propagators will be used.
> Wherever Feynman diagrams will have internal lines and loops,
> integration over the momenta of these internal lines is still required.
> These internal lines whose momenta must be integrated over are commonly
> referred to as "virtual particles". In this formulation, there are no
> bare particles, but still there are virtual particles. Conclusion: bare
> particles have nothing to do with virtual particles and vice versa.
> And, I'm afraid I'm repeating myself, "virtual particles" are not
> particles at all!

I think we are moving closer to agreement here.
If you are interested only in the S-matrix, you have a great freedom
in choosing the Hamiltonian, particle representation, and Feynman rules.
you can use the bare particle representation + renormalization (as it
was done more that 50 years ago) and get exact results. However, if you
are interested in the time evolution of wave functions, you MUST choose
the Hamiltonian in the dressed particle representation. Otherwise,
you may easily get a wrong physical interpretation. Otherwise, in your
theory a single electron will dissociate into a bunch of multiparticle
states, and you may get a wrong impression that two electrons interact
with each other by throwing virtual particles.

This never happens in the dressed particle representation. A single
electron remains single at all times. Two electrons interact via
instantaneous potentials like V = a^+a^+aa. Of course, when the
S-matrix is calculated with these potentials you'll meet products
like V^2, V^3, etc. which can be represented by loop diagrams in the
Feynman's technique.

The point we were discussing was how real particles interact.
The answer is not obvious unless you express the interaction operator
in terms of creation and annihilation operators of real (dressed)
particles. If you do that, then the instantaneous (not retarded) form
of interaction V = a^+a^+aa becomes obvious.

>
>
>>According to classical electromagnetic theory, electromagnetic waves
>>(= real photons in my language) can be created only by charges in
>>accelerated motion. Now, let us take a van-der-Graaf generator with
>> two
>>highly charged metallic balls. There is no movement of charges
>>accelerated or otherwise. So, there are no electromagnetic waves in
>> this
>>case. If you put the generator in a dark room, place photographic
>>plates everywhere around it, and wait for 100 years, you'll not find
>> a
>>single blackened spot on any of the plates. So, there was no single
>>real photon in the room. However, there was a huge electrostatic
>>attraction between the two
>>balls. How are you going to apply your statement "Electromagnetic
>> waves
>>== electromagnetic interaction" in this case?
>
>
> Yes, I will apply that statement. And if you will indeed perform the
> experminet with a van der Graaf generator, you will see spots on the
> plates. Well, perhaps the photographic paper is not sensitive for low
> enough fequencies, in that case I suggest a simple radio reciever. You
> say that electromagnetic radiation is associated with accelerating
> charges. But tell me, how do you charge up a van der Graaf generator
> without accelerating any charges? Even if you assume that the electric
> fields have settled into a static configuration, the statement that
> this interaction can still be explained with electromagnetic waves is
> nothing but the statement that the electric field can be written in a
> Fourier decomposition in terms of different frequency components.

I am sure that after all moving parts in the generator have stopped
and static field has settled down, you will not be able to detect
neither visible nor radio waves (photons) around it. There could be
some radiation due to tunneling of electrons between the balls,
I am not sure about that. But let us disregard this exotic possibility.

>
> If you've ever witnessed interference in radio or TV signals, have you
> ever wondered why it is called "static"?

That's wrong usage of the word. There could be no EM waves emitted
by a static set of charges in the ground state.

>
> And if you think of the thought experiment that I propose to you (take
> to relatively stationary particles, perturb one of them at event (x,t)
> and measure the event (x',t') where the second particle will start
> feeling the effects) you'll see that particle one cannot be perturbed
> without suffering acceleration and thus emitting radiation. There is
> nothing mysterious that the space-time separation between events (x,t)
> and (x',t') can only be light-like.

I disagree. According to my interpretation of dressed particle QED,
the second particle at (x',t) will start to feel the perturbation of
the first particle (x,t) immediately (t'=t) due to the direct Coulomb
and magnetic a^+a^+aa terms in the Hamiltonian. There will be another
kick felt by the second particle at time t' = t + |x' - x|/c
due to REAL photons emitted by the accelerated 1st particle.
This second kick is what is usually described by the transverse
electromagnetic
wave. The first kick (at t'=t) is due to action-at-a-distance.

So far there were no experiments unambiguously confirming/rejecting
my proposal. The closest thing is the observation of superluminal
"photon tunneling" in FTIR. However, I agree that this experiment
allows different interpretations. Let's wait and see when
experimentalists will come up with something better.

>
>
>>As I understand, you think that the Coulomb interaction was
>>transmitted by some material carrier (field) which cannot be detected
>>by itself (it doesn't blacken photographic plates, it doesn't leave
>>traces in bubble chambers, etc). In other words, the field is a form
>>of matter quite different from real particles (photons, electrons,
>>neutrinos, etc). I am not talking about the transverse
>> electromagnetic
>>field which freely propagates in space and, in my language, is
>>equivalent to the flow of real particles photons. I am talking about
>>the static Coulomb field. The only way to detect such a field is by
>> its
>>action on charged particles.
>
>
> Blackening photographic plates is not the only way to detect a field.
> Your eyes, your radio, and charged particles in the ionosphere do a
> good job as well. The electromagnetic field is detectable and changes
> in the electromagnetic field and the field itself *are* electromagnetic
> waves, that is can they can be decomposed in oscillating Fourier modes.

Eyes and radio detect transverse EM fields, that I prefer to call
particles (photons). In RQD they are equivalent to freely propagating
photons. Static (Coulomb and magnetic) fields can only
be detected by forces exerted on test charges. In RQD, these forces are
described by the terms like a^+a^+aa in the Hamiltonian.
These forces are instantaneous (the force acting on particle A depends
on the position and velocity of particle B at the same time instant).
 From this I conclude that EM interaction is a combination of
action-at-a-distance and "exchange" of real photons moving at the
speed of light.

>
>
>>My interpretation is that there is no such field. The interactions
>>between charged particles can be described more economically by
>>using direct instantaneous interparticle potentials
>>(Coulomb and magnetic). I understand that currently there is no
>>convincing experiment that allows us to distinguish between our two
>>interpretations. So, I admit that your point of view has a right to
>>exist. All I am asking from you is to admit that my point of view
>>does not
>>violate any sacred physical principle, does not contradict
>> experiment,
>>and also has a right for existence. Let future experiments decide who
>>was right.
>
>
> Your interpretation is at odds with not only with the conventional one,
> but from the above I can also see that you do not recognize well
> established experimental evidence. So far I have not seen you propose
> an experiment that would be able to falsify your hypothesis. Care to
> speculate?

The idea of the experiment is simple (its practical implementation is not):
as you said, take 2 charges at some distance, wiggle the first charge,
and record the time when the second charge starts to wiggle in response.

As I discuss in subsection 12.3.5 of the book, the FTIR experiment
is very close to this ideal: when the light wave or microwave falls on
the glass-air interface it reflects and induces vibration of dipoles
on the interface. This creates time-varying Coulomb and magnetic fields
in the gap. They are usually called "evanescent waves". It is important
that the transverse field is not present in the gap (it totally
reflects), so there is no admixture of the retarded interactions to
the forces acting between dipoles on two sides of the gap. According
to my views, these forces are purely instantaneous. The induced
oscillation of dipoles on the other side of the gap creates EM wave
(photons) propagating in the other piece of glass. This makes an
impression that photons have tunnelled across the gap faster than
light.

I thought about changing this experiment by eliminating the transverse
fields entirely. For example, one can use ultrasound waves instead
of microwaves. If one can generate a wave of optical phonons
(polarization) in the material, this wave will also reflect from the
interface and create time-varying dipole moments there. These dipoles
will induce dipoles on the other side of the gap, and a corresponding
ultrasound wave there. If I am right, then the dipoles on both
sides of the gap interact instantaneously, and the whole process
will be seen as superluminal tunneling of ultrasound across the gap.
Unfortunately the period of ultrasound vibrations is many orders
of magnitude larger that the time required for light to cross the
gap. So, it is nearly impossible to distinguish superluminal and
subluminal situations in such experiment.

I invite you to give some thought to possible experimental
tests of "retarded vs. instantaneous" interactions. In my opinion,
such an experiment is long overdue.

Eugene Stefanovich.

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